if [tex] g(s)= \sum_{n=1}^{\infty} a(n) n^{-s} [/tex] Where g(s) has a single pole at s=1 with residue C, then my question/conjecture is if for s >0 (real part of s bigger than 0) we can write [tex] g(s)= C(\frac{1}{s-1}+1)-s\int_{0}^{\infty}dx(Cx-A(x))x^{-s-1} [/tex] of course [tex] A(x)=\sum_{n \le x}a(n) [/tex] the question is if the series converge for s >1 with a pole there is a method to 'substract' this singularity (pole) at s=1 to give meaning for the series at any positive s. I think that the 'Ramanujan resummation' may help to give the result: [tex] \sum_{ n >1}^{[R]}a(n)n^{-s} = g(s)-C(s-1)^{-1} [/tex] valid even for s=1 or s>0 (??)